Question

Chứng minh A = \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{100^2}\)< \(\frac{1}{4}\)

Answer 1

A=1/4^2+1/6^2+1/8^2+....+1/100^2=1/16+1/36+1/64+...+1/10000<1/15+1/35+1/63+...+1/9999

Ta có :1/15+1/35+1/63+...+1/9999=1/3.5+1/5.7+1/7.9+...+1/99.101=1/2.(2/3.5+2/5.7+2/7.9+...+2/99.101)

=1/2.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/99-1/101)=1/2.(1/3-1/101)=1/2.100/101=100/202=50/101

Answer 2

bạn Hoàng ơi ban giải bị sai rồi kìa